$\begin{aligned} &f(x)=3\cdot 2^{x-4} \\\\ &g(x)=\dfrac{1}{1+x} \end{aligned}$ $g(f(2))=$
When evaluating composite functions, we work our way inside out. To evaluate $g(f(2))$, let's first evaluate $f(2)$. Then we'll plug that result into $g$ to find our answer. Let's evaluate $f({2})$. $\begin{aligned}f(x)&=3\cdot 2^{x-4}\\\\ f({2})&=3\cdot 2^{({2})-4} ~~~~~~~~~~\text{Plug in }x={2}\\\\ &=3\cdot 2^{-2}\\\\ &=3\cdot \dfrac{1}{4}\\\\ &={\dfrac{3}{4}}\end{aligned}$ We now know that $g(f({2}))$ is the same as $g\left({\dfrac{3}{4}}\right)$ because $f({2}) = {\dfrac{3}{4}}$. Let's evaluate $g\left({\dfrac34}\right)$. $\begin{aligned}g(x)&=\dfrac{1}{1+x}\\\\ g\left({{\dfrac34}}\right)&=\dfrac{1}{1+\left({\dfrac{3}{4}}\right)}~~~~~~~~~~\text{Plug in }x={\dfrac{3}{4}}\\\\ &=\dfrac{1}{\dfrac{7}{4}}\\\\\\ &=\dfrac{4}{7}\end{aligned}$ The answer: $g(f(2)) = \dfrac{4}{7}$